LCP 06.Coin

字数统计: 65字   |   阅读时长: 1min

topic：

screenshot2023-09-21 morning12.12.09.png
topic链接

Thought：

now that4Two times，3也Two times，So it means that this question is nothing more than“As much as possible2indivual2indivual取，Turn count+1”
Although it can be optimized at running time，But I prefer to write a line

Code：

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class Solution:
    def minCount(self, coins: List[int]) -> int:
        return sum(i // 2 + i % 2 for i in coins)

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2085. Public string that has occurred once

字数统计: 95字   |   阅读时长: 1min

topic：

screenshot2024-01-12 afternoon5.12.50.png

Thought：

First useCountercount，use & The symbol is connected to two dictionarieskeys， The remaining part is both。
Traversing this remaining part，Determine whether it is equal to 1 Be

Code：

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class Solution:
    def countWords(self, words1: List[str], words2: List[str]) -> int:
        words_1 = Counter(words1)
        words_2 = Counter(words2)
        ans = 0
        for i in words_1.keys() & words_2.keys():
            if words_1[i] == 1 and words_2[i] == 1:
                ans += 1
        return ans
One -line writing
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return sum(1 for i in Counter(words1) & Counter(words2) if Counter(words1)[i] == 1 and Counter(words2)[i] == 1)

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1004.Maximum continuity1Number III Maximum continuity1Number III

字数统计: 297字   |   阅读时长: 1min

Today’s daily question is too difficult，So find the problem by yourself。Today’s question is a hash table+Sliding window algorithm，虽然感觉只用了Sliding window algorithm。

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class Solution:
    def longestOnes(self, nums: List[int], k: int) -> int:
        k_mean = k
        # Used to record how many arrays there are now
        flag = 0
        # Used to record the maximum land number
        max_flag = 0
        for start in range(len(nums)):
            tail = start
            while k >= 0 and tail <= len(nums) - 1:
                if nums[tail] == 1:
                    tail += 1
                    flag += 1
                elif nums[tail] == 0 and k > 0:
                    tail += 1
                    k -= 1
                    flag += 1
                elif nums[tail] == 0 and k == 0:
                    k = k_mean
                    max_flag = max(max_flag, flag)
                    flag = 0
                    break
                if tail == len(nums):
                    max_flag = max(max_flag, flag)
                    flag = 0
                    break
        return max_flag

This is my approach at the beginning，Although the double pointer is used，But there is no flexibility，Very empty feeling，Very dry slide。
the following@Lincoln@Lincoln Big practice，Just as one record of yourself，不作为我的answer发表

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class Solution:
    def longestOnes(self, nums: List[int], k: int) -> int:
        """
        Thinking：1. k=0It can be understood as the problem of the maximum duplication sub -string
             2. If the current window value-In the window1Number <= k: Then expand the window(right+1)
                If the current window value-In the window1Number > k: Swipe the window to the right(left+1)
        method：Hash table + Sliding window
        """
        n = len(nums)
        o_res = 0
        left = right = 0
        while right < n:
            if nums[right]== 1: o_res += 1
            if right-left+1- o_res > k:
                if nums[left]== 1: o_res -= 1 
                left += 1
            right += 1
        return right - left

Look atLincolnBigThinking，very clearly，remember

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